Question

0.768 gm of salt Na₂X is added to water to prepare 50 ml aq. solution. The solution is titrated with V ml 0.1 M H₂SO₄ aq. solution. If molar mass of weak acid 'H₂X' is 52 gm mol⁻¹ & [Ka₁(H₂X) = 10⁻⁵, Ka₂(H₂X) = 10⁻⁸]. Identify correct option(s) for the solution formed. [Given : log2 = 0.3, log3 = 0.48]

• A. If V = 20 then pH of final solution is 8
• B. If V = 50 then pH of final solution is 5.48
• C. If V = 80 then pH of final solution is 9.52
• D. If V = 100 then pH of final solution is nearly 1.58

Answer 1

Answer: (A), (B), (D)

Options (A), (B), (D) are correct.

Answer 2

We start with 0.768 g of Na₂X. The molar mass of H₂X is 52 g/mol so that the X part has mass 52 – 2 = 50 g/mol. Hence, molar mass of Na₂X is 2×23 + 50 = 96 g/mol and moles of Na₂X = 0.768/96 = 0.008 mol. In 50 mL the [X²⁻] = 0.008/0.05 = 0.16 M.

Since Na₂X comes from the diprotic acid H₂X (with Ka₁ = 10⁻⁵, pKa₁ = 5 and Ka₂ = 10⁻⁸, pKa₂ = 8), titrating X²⁻ with acid occurs in two steps:

1. X²⁻ + H⁺ → HX⁻  (using pKa₂ = 8)
2. HX⁻ + H⁺ → H₂X  (using pKa₁ = 5)

Recall that 0.1 M H₂SO₄ provides 2 moles of H⁺ per mole so the equivalents of acid added = 0.0002 × V (with V in mL).

Define the following regions:

─────────────────────────────

Step 1 complete when: 0.0002V = 0.008 → V = 40 mL.

─────────────────────────────

Now analyze each option:

(A) V = 20 mL:

Acid equivalents = 0.0002×20 = 0.004.

Reaction:

• X²⁻ decreases: 0.008 – 0.004 = 0.004 mol
• HX⁻ is formed: 0.004 mol

Total volume = 50 + 20 = 70 mL

Buffer: HX⁻ (acid) and X²⁻ (base) with pKa₂ = 8.

Using Henderson–Hasselbalch:

pH = 8 + log([X²⁻]/[HX⁻]) = 8 + log(0.004/0.004) = 8 + 0 = 8.

Option (A) is correct.

(B) V = 50 mL:

Acid equivalents = 0.0002×50 = 0.01.

First, 0.008 mol of X²⁻ is completely converted to HX⁻;

Excess acid = 0.01 – 0.008 = 0.002 mol converts HX⁻ to H₂X.

So,

HX⁻ remaining = 0.008 – 0.002 = 0.006 mol,

H₂X formed = 0.002 mol.

Total volume = 50 + 50 = 100 mL

Now the buffer is the HX⁻/H₂X system (pKa₁ = 5):

pH = 5 + log([HX⁻]/[H₂X]) = 5 + log(0.006/0.002) = 5 + log(3).

Given log 3 = 0.48, hence pH = 5.48.

Option (B) is correct.

(C) V = 80 mL:

Acid equivalents = 0.0002×80 = 0.016.

First 0.008 mol acid neutralizes X²⁻ to HX⁻, then the next 0.008 mol converts HX⁻ completely to H₂X.

We have only H₂X left (0.008 mol).

Total volume = 50 + 80 = 130 mL → [H₂X] = 0.008/0.13 ≈ 0.0615 M.

For a weak acid,

pH ≈ ½(pKa – log C) = ½(5 – log(0.0615)).

Since log(0.0615) ≈ log(6.15×10⁻²) = log6.15 – 2.

Approximating log6.15 ≈ 0.79 gives log(0.0615) ≈ -1.21,

so pH ≈ ½(5 + 1.21) ≈ 3.105.

Option (C) stating pH 9.52 is incorrect.

(D) V = 100 mL:

Acid equivalents = 0.0002×100 = 0.02.

Neutralization steps:

• X²⁻ (0.008 mol) → HX⁻;
• HX⁻ (0.008 mol) → H₂X;

Total acid consumed for neutralization = 0.016 mol.

Excess acid = 0.02 – 0.016 = 0.004 mol free H⁺.

Total volume = 50 + 100 = 150 mL

H⁺ concentration = 0.004/0.15 ≈ 0.02667 M,

pH = –log(0.02667) ≈ 1.58.

Option (D) is correct.